Doping & PN Junctions

Question 1

(a) (5 marks) In equilibrium the Fermi level is flat. On p-side $p\approx N_A$; on n-side $n\approx N_D$. Using $np=n_i^2$:

• p-side hole density $= N_A$, n-side hole density $=n_i^2/N_D$. (1)
• Built-in potential balances diffusion; using $p(x)=p_0 e^{-qV/k_BT}$ over the junction: $\frac{p_p}{p_n}=e^{qV_{bi}/k_BT}=\frac{N_A}{n_i^2/N_D}=\frac{N_A N_D}{n_i^2}$ (2)
• Solving: $V_{bi}=\dfrac{k_BT}{q}\ln\!\dfrac{N_AN_D}{n_i^2}$. (1)

Numerics: $V_T=0.02585\ \text{V}$; argument $=\frac{10^{17}\cdot10^{16}}{(10^{10})^2}=10^{13}$; $V_{bi}=0.02585\times\ln(10^{13})=0.02585\times29.93=0.774\ \text{V}$. (1)

(b) (7 marks) Poisson (1-D): $\dfrac{dE}{dx}=\dfrac{\rho}{\varepsilon}$, with $\rho=-qN_A$ ($-x_p<x<0$), $\rho=+qN_D$ ($0<x<x_n$). (1) Integrate each region; field is triangular, peaking at $x=0$ and zero at edges:

• $E_{max}=qN_A x_p/\varepsilon = qN_D x_n/\varepsilon$ ⇒ charge neutrality $N_A x_p=N_D x_n$ (condition 1). (2)
• Continuity of $E$ at $x=0$ used above; $E=0$ at depletion edges (condition 2). (1) Potential $=\int E\,dx$ = area of triangle: $V_{bi}=\tfrac12 E_{max}(x_p+x_n)=\tfrac12 E_{max}W$. (1) With $x_p=\frac{N_D}{N_A+N_D}W$, $x_n=\frac{N_A}{N_A+N_D}W$ and $E_{max}=\frac{q}{\varepsilon}\frac{N_AN_D}{N_A+N_D}W$: $V_{bi}=\frac{q}{2\varepsilon}\frac{N_AN_D}{N_A+N_D}W^2\ \Rightarrow\ W=\sqrt{\frac{2\varepsilon}{q}\Big(\tfrac1{N_A}+\tfrac1{N_D}\Big)V_{bi}}.$ (2)

(c) (5 marks) $\varepsilon=11.7\times8.854\times10^{-12}=1.036\times10^{-10}\ \text{F/m}$. Convert doping to m$^{-3}$: $N_A=10^{23}$, $N_D=10^{22}$. $\frac1{N_A}+\frac1{N_D}=10^{-23}+10^{-22}=1.1\times10^{-22}\ \text{m}^3$. (1) $W=\sqrt{\frac{2(1.036\times10^{-10})}{1.602\times10^{-19}}(1.1\times10^{-22})(0.774)}$ $=\sqrt{2\cdot1.036\times10^{-10}/1.602\times10^{-19}\cdot0.774\cdot1.1\times10^{-22}}$. Inner: $2\cdot1.036\times10^{-10}=2.072\times10^{-10}$; $/1.602\times10^{-19}=1.293\times10^{9}$; $\times0.774=1.001\times10^{9}$; $\times1.1\times10^{-22}=1.101\times10^{-13}$; $W=\sqrt{1.101\times10^{-13}}=3.32\times10^{-7}\ \text{m}=0.332\ \mu\text{m}$. (2) Fraction in n-side $=x_n/W=\frac{N_A}{N_A+N_D}=\frac{10^{17}}{1.1\times10^{17}}=0.909$ (≈91 %). (1) Physical: to expose equal net charge each side ($N_Ax_p=N_Dx_n$), the lightly-doped n-side (fewer dopants/vol) must deplete a wider region. (1)

(d) (5 marks) $E_{max}=2V_{bi}/W=2\times0.774/3.32\times10^{-7}=4.66\times10^{6}\ \text{V/m}=46.6\ \text{kV/cm}$. (2) Sketch relationships (3):

• $\rho(x)$: step: $-qN_A$ (p-side), $+qN_D$ (n-side), narrower on p-side.
• $E(x)$: triangular, negative dip, peak magnitude at $x=0$; $E=-\frac1\varepsilon\int\rho$… i.e. $dE/dx=\rho/\varepsilon$.
• $V(x)$: smooth S-curve rising by $V_{bi}$; $E=-dV/dx$, so $V=-\int E\,dx$.

Question 2

(a) (5) $V_T=k_BT/q=1.381\times10^{-23}\times300/1.602\times10^{-19}=0.02585\ \text{V}$. (2) For $V\gg V_T$: $I\approx I_S e^{V/V_T}$. Per decade: $10=e^{\Delta V/V_T}$ ⇒ $\Delta V=V_T\ln10=0.02585\times2.3026=0.0595\ \text{V}\approx60\ \text{mV}$. (3)

(b) (5) $V=V_T\ln(I/I_S+1)\approx V_T\ln(10^{-3}/10^{-14})=0.02585\times\ln(10^{11})=0.02585\times25.33=0.655\ \text{V}$. (3) $dI/dV=I/V_T$ (large forward), so $r_d=V_T/I=0.02585/10^{-3}=25.85\ \Omega$. (2)

(c) (7) Iteration: solve $f(V)=I_S(e^{V/V_T}-1)-I_{target}=0$, $f'(V)=(I_S/V_T)e^{V/V_T}$. $V_{k+1}=V_k-\frac{I_S(e^{V_k/V_T}-1)-I_{target}}{(I_S/V_T)e^{V_k/V_T}}$

import math
def solve_diode_V(Is, T, Itarget, tol=1e-12, itmax=100):
    kB = 1.381e-23; q = 1.602e-19
    VT = kB*T/q
    V = VT*math.log(Itarget/Is + 1.0)   # good initial guess (ideal-diode inverse)
    for _ in range(itmax):
        f  = Is*(math.exp(V/VT) - 1.0) - Itarget
        fp = (Is/VT)*math.exp(V/VT)
        step = f/fp
        V -= step
        if abs(step) < tol:             # stopping criterion
            break
    return V

Marks: correct $f,f'$ (2); iteration formula (2); sensible guess (1); stopping criterion (1); runnable structure (1).

(d) (5) $I_S\propto T^3 e^{-E_g/k_BT}$; the exponential dominates. $\frac{d\ln I_S}{dT}=\frac{3}{T}+\frac{E_g}{k_BT^2}$; for Si ($E_g\approx1.12$ eV) at 300 K the second term $\gg$ first, giving ≈0.08/°C ⇒ $e^{0.08\times10}\approx2.2$, hence roughly doubling per 10 °C. (3) Consequence: at fixed forward $V$, $I=I_S e^{V/V_T}$ rises with $T$, risking thermal runaway (self-heating raises $I_S$, raises current, raises heating). (2)

Question 3

(a) (5) $W\propto(V_{bi}+V_R)^{1/2}$ (from Q1b with $V_{bi}\to V_{bi}+V_R$). $C_j=\varepsilon/W\propto(V_{bi}+V_R)^{-1/2}$, so $m=1/2$ for an abrupt junction. (3) Justification: reverse bias widens $W$, separating the plates further, lowering capacitance; the $\sqrt{}$ arises from the linear-field/Poisson integration. (Linearly-graded junctions give $m=1/3$.) (2)

(b) (6)

• Avalanche: carriers accelerated by high field gain enough energy to impact-ionise, multiplying carriers; occurs in lightly-doped, wide junctions (higher $V_{BD}$, >~6 V). Positive temperature coefficient — hotter lattice → more phonon scattering → carriers reach ionisation energy over longer distance → higher $V_{BD}$. (3)
• Zener: direct field-induced band-to-band tunnelling across a very narrow depletion region; occurs in heavily-doped junctions (thin $W$, low $V_{BD}$, <~5 V). Negative temperature coefficient — higher $T$ narrows $E_g$, easing tunnelling → lower $V_{BD}$. (3)

(c) (5) $5.6\ \text{V}$ lies in the transition band but the positive temperature coefficient is decisive: it indicates avalanche (Zener has negative coefficient). Above ~6 V avalanche dominates; the positive sign confirms impact-ionisation is the mechanism here. (5)

[
  {"claim":"V_bi = 0.774 V for NA=1e17, ND=1e16, ni=1e10 at 300K",
   "code":"VT=1.381e-23*300/1.602e-19; Vbi=VT*log(1e17*1e16/(1e10)**2); result=abs(Vbi-0.774)<0.01"},
  {"claim":"Depletion width W approx 0.332 um",
   "code":"import sympy as sp; eps=11.7*8.854e-12; q=1.602e-19; VT=1.381e-23*300/q; Vbi=VT*sp.log(1e17*1e16/(1e10)**2); NA=1e23; ND=1e22; W=sp.sqrt(2*eps/q*(1/NA+1/ND)*Vbi); result=abs(float(W)-3.32e-7)<0.1e-7"},
  {"claim":"Fraction of W in n-side = 0.909",
   "code":"NA=1e17; ND=1e16; frac=NA/(NA+ND); result=abs(frac-0.909)<0.01"},
  {"claim":"Forward voltage for 1mA with Is=1e-14 is about 0.655 V",
   "code":"VT=1.381e-23*300/1.602e-19; V=VT*log(1e-3/1e-14+1); result=abs(V-0.655)<0.005"},
  {"claim":"Dynamic resistance rd = 25.85 ohm at 1mA",
   "code":"VT=1.381e-23*300/1.602e-19; rd=VT/1e-3; result=abs(rd-25.85)<0.5"},
  {"claim":"Voltage per decade approx 59.5 mV",
   "code":"VT=1.381e-23*300/1.602e-19; dV=VT*log(10); result=abs(dV-0.0595)<0.001"}
]